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Inquiry Question 3: It is all about hydrogen ions

Calculate concentration changes on dilution using c1v1 = c2v2 and predict the effect of dilution on pH for strong and weak acid and base solutions

A focused answer to the HSC Chemistry Module 6 dot point on dilution. Concentration units (mol/L, percent w/v, ppm), the dilution equation c1v1 = c2v2, how pH changes on dilution for strong and weak acids, and worked HSC past exam questions.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
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In plain English

Think of a concentrated acid as cordial straight from the bottle, and diluting it as topping the glass up with water. The number of "cordial molecules" (moles of solute) in the glass never changes, only the volume of liquid they are spread through, so a bigger glass of the same cordial always tastes weaker in direct proportion, that is exactly what $c_1V_1 = c_2V_2$ says. But acids come in two personalities. A strong acid is like a cordial that always releases all of its flavour into the water at once. Halve its concentration and the taste halves exactly, so its pH climbs by a clean, predictable one unit every time you make a ten times bigger glass. A weak acid is more like a tea bag that only releases some of its flavour while it sits in the cup. Dunk that same tea bag in a much bigger cup of water and, strangely, it releases a slightly higher fraction of its flavour into the water than it did in the small cup, this is Le Chatelier's principle at work on the ionisation equilibrium, so the taste (and the pH change) does not fall as much as you would expect from the extra water alone. That is why a weak acid's pH rises by only about half a unit, not a full unit, for the same ten-fold dilution.

What this dot point is asking

NESA wants you to handle the quantitative side of acid-base chemistry: convert between concentration units, perform serial and one-step dilutions, and predict how pH changes when an acid or base is diluted. The mathematics is the dilution identity $c_1 V_1 = c_2 V_2$ ; the chemistry is the difference between strong and weak acids on dilution, which builds on the strong vs weak page and feeds into titration analysis.

The answer

Concentration units

Unit	Symbol	Meaning
Molarity	mol/L (M)	moles of solute per litre of solution
Mass percent	percent w/w	g solute per 100 g solution
Mass/volume percent	percent w/v	g solute per 100 mL solution
Volume percent	percent v/v	mL solute per 100 mL solution
Parts per million	ppm	mg solute per L solution (for dilute aqueous solutions)
Parts per billion	ppb	$\mu$ g solute per L solution

Molarity is the default HSC unit because it links directly to stoichiometry ( $n = cV$ ). The other units appear in environmental and biological contexts (lead in drinking water, residual chlorine, salinity).

Conversions to remember.

$1 \text{ ppm} \approx 1 \text{ mg/L}$ for dilute aqueous solutions (because 1 L of water has mass 1 kg).
$c \text{ (mol/L)} = \frac{\text{percent w/v} \times 10}{M_r}$ , where $M_r$ is the molar mass in g/mol.

The dilution equation

When solvent is added to a solution, the moles of solute do not change. Therefore:

n_1 = n_2 \quad \Rightarrow \quad c_1 V_1 = c_2 V_2

Use any consistent volume unit (mL or L) as long as both sides match. The equation works for any solute, not just acids and bases.

Procedure for an accurate dilution.

Calculate the volume of stock needed: $V_1 = c_2 V_2 / c_1$ .
Transfer $V_1$ to a volumetric flask of size $V_2$ using a pipette (for small volumes) or a measuring cylinder.
Add distilled water to roughly three-quarters full, swirl.
Top up to the calibration mark, with a dropper for the last few drops.
Stopper, invert several times to ensure homogeneity.

pH on dilution: strong acid or base

For a strong acid, $[H^+] = c$ (fully ionised). On dilution, $[H^+]$ drops in direct proportion to $c$ . A 10-fold dilution raises pH by 1.0 unit; a 100-fold dilution raises pH by 2.0 units.

For a strong base, $[OH^-] = c$ on the same logic, and pH falls by 1 unit per 10-fold dilution (because pOH rises by 1).

Limit: as you dilute past about $10^{-6}$ mol/L, the auto-ionisation of water becomes significant and pH approaches 7 from below (for an acid) or above (for a base), never crossing 7. A common HSC error is to claim that $10^{-9}$ mol/L $HCl$ has pH 9; in fact it has pH just under 7 because the dominant source of $H^+$ is now water itself.

pH on dilution: weak acid or base

For a weak acid, $[H^+] \approx \sqrt{K_a c}$ (from the ICE shortcut, valid when ionisation is small).

A 10-fold dilution changes $c$ by 10, so $[H^+]$ changes by $\sqrt{10} \approx 3.16$ , and pH rises by $\log_{10}(\sqrt{10}) = 0.5$ . Weak acids resist pH change on dilution more than strong acids do.

Mechanistically (Le Chatelier): the ionisation equilibrium has more moles of dissolved particles on the right side, so dilution favours further ionisation. The percent ionisation rises as concentration falls, partially offsetting the dilution.

Summary rule. A 10-fold dilution raises pH by:

1.0 unit for a strong acid (until water dominates).
~0.5 unit for a weak acid.
~0 for a buffer (until exhausted).

An owned illustrative titration curve shows how the equivalence-point pH gives away whether the original acid was strong or weak, the reasoning needed for the DATA question later on this page:

Serial dilution

If you need a very dilute solution, a single one-step dilution can be inaccurate (pipetting a very small volume). Serial dilution does it in stages: each step is a 10- or 100-fold dilution. To go from 1.00 mol/L to $1.00 \times 10^{-4}$ mol/L, do four successive 10-fold dilutions.

Worked example

Calculate the pH of (a) $0.0100$ mol/L $HCl$ and (b) $0.0100$ mol/L $CH_3COOH$ ( $K_a = 1.8 \times 10^{-5}$ ), then state the pH that each would have after a 100-fold dilution.

(a) HCl, initial: $[H^+] = 0.0100$ mol/L, $pH = 2.00$ .
(a) HCl after 100-fold dilution: $c = 1.00 \times 10^{-4}$ mol/L, $pH = 4.00$ .
(b) Ethanoic acid, initial

[H^+] \approx \sqrt{(1.8 \times 10^{-5})(0.0100)} = \sqrt{1.8 \times 10^{-7}} = 4.24 \times 10^{-4} \text{ mol/L}

pH = -\log_{10}(4.24 \times 10^{-4}) = 3.37

(b) Ethanoic acid after 100-fold dilution. $c = 1.00 \times 10^{-4}$ mol/L:

[H^+] \approx \sqrt{(1.8 \times 10^{-5})(1.00 \times 10^{-4})} = \sqrt{1.8 \times 10^{-9}} = 4.24 \times 10^{-5} \text{ mol/L}

pH = -\log_{10}(4.24 \times 10^{-5}) = 4.37

The strong acid rose by 2.0 pH units (as predicted: 100-fold dilution, $\log_{10}(100) = 2$ ). The weak acid rose by 1.0 pH unit (as predicted: $\log_{10}(\sqrt{100}) = 1$ ). At this point the weak acid pH (4.37) is higher than the strong acid pH (4.00) because the strong acid is still fully ionised and the weak acid is not, even though both share the same nominal concentration.

Common traps

Forgetting consistent units: In $c_1 V_1 = c_2 V_2$ , both volumes must be in the same unit. Mixing mL and L drops a factor of 1000.
Applying $[H^+] = c$ to a weak acid: That formula is for strong acids only. Use $\sqrt{K_a c}$ for weak acids.
Extrapolating strong-acid pH past auto-ionisation: $10^{-9}$ mol/L $HCl$ does not have pH 9. Below about $10^{-6}$ mol/L, water's auto-ionisation contributes; a full charge balance gives pH approaching 7 from the acidic side.
Calling dilute strong acid "weak": Dilution lowers concentration; it does not change the strength (degree of ionisation). A dilute strong acid is still a strong acid (still fully ionised).
Confusing percent w/v with mol/L: A 5 percent w/v $NaOH$ solution is 50 g/L, which converts to $50/40 = 1.25$ mol/L. Always carry the conversion through molar mass.
Pipetting from a volumetric flask without rinsing: When making a standard, the pipette should be rinsed with the stock first (not water), and the volumetric flask should be rinsed with distilled water (not stock).

Examples in context

Example 1. Diluting bench acid in the NSW HSC prac lab. A standard exercise has students prepare 250 mL of 0.100 mol L $^{-1}$ HCl from a 2.00 mol L $^{-1}$ bench stock. Rearranging $c_1 V_1 = c_2 V_2$ gives $V_1 = 0.100 \times 0.250 / 2.00 = 12.5$ mL. The student pipettes 12.5 mL of stock into a 250 mL volumetric flask, swirls, and tops up to the mark. NSW DET safety guidelines require acid into water (never the reverse) and use of a safety pipette filler. The resulting solution has pH 1.00 because HCl is fully ionised. The same calculation underpins every clinical IV dilution at NSW Health.

Example 2. Dilution effects on weak versus strong acid in environmental sampling. A WaterNSW officer collects acidic mine drainage near Captains Flat at $[H^+] = 0.010$ mol L $^{-1}$ (pH 2.0) and a sample of dilute acetic acid runoff from a vineyard at 0.010 mol L $^{-1}$ acetic acid ( $K_a = 1.8 \times 10^{-5}$ , pH 3.4). On a 100-fold dilution to $1.0 \times 10^{-4}$ mol L $^{-1}$ , the strong acid drops to pH 4.0 (a rise of 2 units) but the weak acid rises only to about pH 4.7. Officers use this contrast to confirm whether a low pH reading reflects strong mineral acidity or weak organic acid, an evidentiary distinction that affects how runoff is regulated.

Try this

Q1. State the dilution equation $c_1 V_1 = c_2 V_2$ and explain why moles of solute are conserved on dilution. [2 marks]

Cue. Equation conserves moles ( $n = cV$ ); adding solvent does not create or destroy solute particles.

Q2. Calculate the volume of concentrated 18.0 mol L $^{-1}$ $H_2SO_4$ required to prepare 500 mL of 0.250 mol L $^{-1}$ acid. [2 marks]

Cue. $V_1 = c_2 V_2 / c_1 = (0.250 \times 0.500) / 18.0 = 6.94$ mL.

Q3. A 0.10 mol L $^{-1}$ solution of HCl and a 0.10 mol L $^{-1}$ solution of acetic acid (pKa = 4.74) are each diluted 100-fold. (a) Calculate the pH of each before dilution. (b) Calculate the pH of each after dilution. (c) Account for the difference in pH change. [2+2+2 marks]

Cue. (a) HCl pH = 1.0; acetic pH = 2.87. (b) HCl pH = 3.0; acetic pH approx 3.87. (c) Strong acid: pH rises by 2 (full ionisation); weak acid: equilibrium shifts right, pH rises by only 1.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC4 marksA 25.0 mL aliquot of 0.500 mol/L HCl is diluted to a final volume of 250.0 mL with distilled water. Calculate (a) the concentration of the diluted solution and (b) the pH before and after dilution. (c) State, without further calculation, how the pH would have changed if the original acid had been 0.500 mol/L ethanoic acid (Ka = 1.8 x 10^-5).

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(a) Dilution. Apply $c_1V_1 = c_2V_2$ :

c_2 = \frac{c_1 V_1}{V_2} = \frac{0.500 \times 25.0}{250.0} = 0.0500 \text{ mol/L}

(b) pH before and after. HCl is a strong acid, so $[H^+] = c$ .

pH_1 = -\log_{10}(0.500) = 0.301

pH_2 = -\log_{10}(0.0500) = 1.301

A 10-fold dilution raises the pH by exactly 1 unit for a strong acid (because $\log_{10}(10) = 1$ ).

(c) For ethanoic acid. A 10-fold dilution would raise the pH by less than 1 unit. The reason: as the weak acid is diluted, the equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ shifts right (Le Chatelier favours the side with more particles), so the percent ionisation increases. $[H^+]$ falls by a factor smaller than 10, so the pH change is smaller than 1.

Quantitatively, $[H^+] \approx \sqrt{K_a c}$ , so a 10-fold drop in $c$ gives only a $\sqrt{10} \approx 3.2$ -fold drop in $[H^+]$ , a pH rise of about 0.5.

Markers reward (1) correct $c_1V_1 = c_2V_2$ , (2) two correct pH values with the right number of sig figs, (3) recognising the smaller pH change for the weak acid with a Le Chatelier or $\sqrt{K_a c}$ justification.

2017 HSC3 marksA student needs to prepare 500.0 mL of 0.100 mol/L NaOH from a stock 2.00 mol/L NaOH solution. Describe the procedure, including the volume of stock required, the apparatus used, and how the volume is made up accurately.

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Step 1: Volume of stock.

V_1 = \frac{c_2 V_2}{c_1} = \frac{0.100 \times 500.0}{2.00} = 25.0 \text{ mL}

Procedure.

Use a 25.0 mL bulb pipette to transfer 25.0 mL of the stock NaOH into a 500.0 mL volumetric flask.
Add distilled water to about three-quarters full and swirl to mix.
Top up to the calibration mark with distilled water, using a dropper for the final few drops so that the bottom of the meniscus sits on the line at eye level.
Stopper and invert the flask several times to mix thoroughly.

Markers reward (1) the correct volume from $c_1V_1 = c_2V_2$ , (2) named volumetric glassware (bulb pipette and volumetric flask), (3) correct meniscus reading and mixing technique.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksState the dilution equation and identify what quantity is conserved when a solution is diluted with water.

Show worked solution →

Equation. $c_1V_1 = c_2V_2$ .

What is conserved. The moles of solute, $n$ . Adding solvent increases the volume but does not add or remove solute particles, so $n_1 = n_2$ , and substituting $n = cV$ into both sides gives the dilution equation.

Marking criteria: 1 mark for the correct equation, 1 mark for correctly identifying moles of solute as the conserved quantity (not mass of solvent or volume).

foundation3 marksA 20.0 mL sample of 0.400 mol/L

KOH

is diluted with water to a final volume of 200.0 mL. Calculate the concentration of the diluted solution, showing all working with correct units and significant figures.

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Step 1: identify the known values. $c_1 = 0.400$ mol/L, $V_1 = 20.0$ mL, $V_2 = 200.0$ mL.

Step 2: rearrange the dilution equation for $c_2$ .

c_2 = \frac{c_1 V_1}{V_2}

Step 3: substitute (volumes in the same unit, mL, so they cancel correctly).

c_2 = \frac{0.400 \times 20.0}{200.0} = \frac{8.00}{200.0} = 0.0400 \text{ mol/L}

Step 4: check significant figures. The data (0.400, 20.0, 200.0) each carry 3 significant figures, so the answer is reported to 3 significant figures.

c_2 = 0.0400 \text{ mol/L (3 s.f.)}

Marking criteria: 1 mark for correctly rearranging the equation, 1 mark for correct substitution with consistent volume units, 1 mark for the correct final value with units and 3 significant figures.

core4 marksA stock solution of nitric acid is labelled 15.8 mol/L. A student dilutes 8.00 mL of this stock to a final volume of 250.0 mL. Calculate (a) the concentration of the diluted acid to 3 significant figures, and (b) the pH of the diluted solution to 2 decimal places, given that nitric acid is a strong acid.

Show worked solution →

(a) Dilution.

c_2 = \frac{c_1 V_1}{V_2} = \frac{15.8 \times 8.00}{250.0} = \frac{126.4}{250.0} = 0.5056 \text{ mol/L}

Rounding the intermediate is avoided; rounding only the final answer to 3 significant figures (matching the 3 s.f. of 15.8 and 8.00):

c_2 = 0.506 \text{ mol/L}

(b) pH. $HNO_3$ is a strong acid, fully ionised, so $[H^+] = c_2 = 0.506$ mol/L.

pH = -\log_{10}(0.506) = 0.296

pH = 0.30 \text{ (2 d.p.)}

Marking criteria: 1 mark for correct rearrangement and substitution, 1 mark for the concentration to 3 significant figures, 1 mark for correctly using $[H^+] = c$ for a strong acid, 1 mark for the correct pH to 2 decimal places.

core5 marksThe graph below is an owned illustrative titration curve for the addition of 0.100 mol/L NaOH to 25.0 mL of an acid of unknown concentration and strength, with the equivalence point marked at 30.0 mL of NaOH added and pH 8.9 at equivalence. (a) State, with a reason, whether the unknown acid is strong or weak. (b) Calculate the concentration of the unknown acid.

Show worked solution →

(a) Strong or weak. The equivalence point pH is 8.9, above 7. For a monoprotic acid titrated with a strong base, an equivalence pH above 7 occurs because the conjugate base of the acid is a weak base that hydrolyses water, which only happens if the acid itself was weak (a strong acid, e.g. $HCl$ , gives a neutral salt and an equivalence pH of exactly 7).

(b) Concentration of the unknown acid. At equivalence, moles of $NaOH$ added equal moles of the (monoprotic) acid originally present.

n(NaOH) = c \times V = 0.100 \text{ mol/L} \times 0.0300 \text{ L} = 3.00 \times 10^{-3} \text{ mol}

n(\text{acid}) = 3.00 \times 10^{-3} \text{ mol (1:1 ratio)}

c(\text{acid}) = \frac{n}{V} = \frac{3.00 \times 10^{-3} \text{ mol}}{0.0250 \text{ L}} = 0.120 \text{ mol/L}

Marking criteria: 1 mark for identifying the acid as weak, 1 mark for the hydrolysis-based reason (not just "pH is above 7"), 1 mark for moles of NaOH at equivalence, 1 mark for the 1:1 mole ratio to the acid, 1 mark for the final concentration with correct units and significant figures.

core4 marksExplain, using Le Chatelier's principle and the equilibrium

CH_3COOH \rightleftharpoons CH_3COO^- + H^+

, why the pH of a weak acid rises by less than 1.0 unit when it is diluted 10-fold, whereas a strong acid of the same initial concentration rises by exactly 1.0 unit.

Show worked solution →

Strong acid: $HCl$ is fully ionised, so $[H^+] = c$ exactly. A 10-fold drop in $c$ gives an exact 10-fold drop in $[H^+]$ , and $pH = -\log_{10}[H^+]$ rises by exactly $\log_{10}(10) = 1.0$ unit.
Weak acid: For $CH_3COOH$ , only a small fraction ionises at equilibrium. Adding water lowers the concentration of all species, but the ionisation equilibrium has more particles (two ions) on the product side than the reactant side (one molecule). By Le Chatelier's principle, diluting the system shifts the equilibrium to the right, increasing the percentage ionisation. This partially replaces some of the $H^+$ that would otherwise have been lost by dilution alone.
Net effect: $[H^+]$ still falls, but by a factor smaller than 10 (specifically $\sqrt{10}$ using $[H^+] \approx \sqrt{K_a c}$ ), so the pH rise is smaller than 1.0 unit, around 0.5 units.

Marking criteria: 1 mark for correctly stating $[H^+] = c$ gives an exact pH rise of 1.0 for the strong acid, 1 mark for identifying the ionisation equilibrium has more particles on the product side, 1 mark for correctly applying Le Chatelier to explain the equilibrium shift on dilution, 1 mark for linking this shift to a smaller pH change than the strong acid.

exam6 marksA laboratory technician must prepare 1.00 L of 0.0200 mol/L

H_2SO_4

from a concentrated stock of 98.0 percent w/w

H_2SO_4

with density 1.84 g/mL. (a) Calculate the molarity of the concentrated stock. (b) Calculate the volume of concentrated stock required. (c) Outline the correct laboratory procedure and safety precautions for this dilution.

Show worked solution →

This is a 6-mark question combining a percent w/w to mol/L conversion, a dilution calculation, and a procedural/safety outline.

(a) Molarity of the concentrated stock. Consider 1.00 L (1000 mL) of stock solution.

m(\text{solution}) = V \times \rho = 1000 \text{ mL} \times 1.84 \text{ g/mL} = 1840 \text{ g}

m(H_2SO_4) = 1840 \text{ g} \times 0.980 = 1803 \text{ g}

n(H_2SO_4) = \frac{m}{M} = \frac{1803 \text{ g}}{98.09 \text{ g/mol}} = 18.39 \text{ mol}

c_1 = \frac{n}{V} = \frac{18.39 \text{ mol}}{1.00 \text{ L}} = 18.4 \text{ mol/L (3 s.f.)}

(b) Volume of stock required. Using $c_1V_1 = c_2V_2$ with $c_2 = 0.0200$ mol/L and $V_2 = 1000$ mL:

V_1 = \frac{c_2 V_2}{c_1} = \frac{0.0200 \times 1000}{18.4} = 1.087 \text{ mL}

V_1 = 1.09 \text{ mL (3 s.f.)}

(c) Procedure and safety.

Add a small volume of distilled water to a 1.00 L volumetric flask first.
Using a well-rinsed graduated (or micro-) pipette and safety filler, measure 1.09 mL of concentrated $H_2SO_4$ and add it slowly to the water in the flask, swirling constantly (acid added to water, never water to concentrated acid, because the dilution is highly exothermic and can spit if reversed).
Allow the flask to cool to room temperature (concentrated $H_2SO_4$ dilution releases significant heat).
Add distilled water to about three-quarters full, then make up to the calibration mark with a dropper for the final drops, reading the meniscus at eye level.
Stopper and invert to mix. Wear safety glasses and gloves throughout, and work in a fume-free, well-ventilated area given the corrosive stock.

Marking criteria: (a) 1 mark for correct mass of solution from density, 1 mark for correct mass of $H_2SO_4$ from percent w/w, 1 mark for the correct stock molarity with units. (b) 1 mark for correctly rearranging and substituting into $c_1V_1 = c_2V_2$ , 1 mark for the correct volume to 3 significant figures. (c) 1 mark for "acid to water" order with a stated reason (exothermic/spitting hazard); a response describing only glassware without the safety order does not earn this mark.

exam7 marksEvaluate the claim that 'dilution always makes an acidic solution safer and less reactive', using the concepts of concentration, strength, and pH developed in this dot point, and referring to both a strong acid and a weak acid example.

Show worked solution →

This is a 7-mark EVALUATE: markers reward a judgement supported by chemistry, not a one-sided description.

Band 6 PLAN.

Define terms precisely: concentration (mol/L) is not the same as strength (degree of ionisation); pH measures $[H^+]$ , a consequence of both.
Case 1 (strong acid): dilution reliably lowers $[H^+]$ in direct proportion to $c$ , raising pH by exactly 1 unit per 10-fold dilution, until water's auto-ionisation dominates near neutral pH. So for a strong acid, dilution genuinely and predictably reduces acidity and hazard, down to a floor near pH 7.
Case 2 (weak acid): dilution also raises pH, but by less per dilution step (about 0.5 units per 10-fold step) because the ionisation equilibrium shifts right (Le Chatelier), increasing percent ionisation. The solution becomes less acidic overall, but a naive reader might wrongly assume "weak acid, already less reactive, dilution barely matters" - in fact the percentage ionisation genuinely increases, an important nuance for the claim.
Complication: dilution changes concentration, not the intrinsic strength/nature of the acid. A very dilute strong acid is still classified as strong (fully ionised) and, if further concentrated again, would behave identically to the original stock. "Safer" claims about corrosivity relate mostly to concentration for direct contact hazards (e.g. spitting, exothermic effects), not only to pH.
Judgement: the claim is broadly true for pH-based reactivity (both acid types become less acidic on dilution) but is an oversimplification because (1) it conflates concentration with strength, (2) it ignores that percent ionisation of a weak acid rises on dilution, and (3) "safer" is not just about pH - dilution reduces corrosive/exothermic hazards but a very dilute strong acid is chemically still the same substance.

Model paragraph (excerpt). The claim that dilution always makes an acidic solution safer holds for the pH-based measure of acidity: diluting a strong acid such as hydrochloric acid lowers $[H^+]$ in direct proportion to concentration, raising pH by exactly one unit per 10-fold dilution until the solution approaches neutral. However, the claim oversimplifies weak-acid behaviour: diluting ethanoic acid also raises its pH, but by only around 0.5 units per 10-fold step, because the equilibrium $CH_3COOH \\rightleftharpoons CH_3COO^- + H^+$ shifts toward greater ionisation as concentration falls, partially offsetting the intended reduction in acidity. The claim is therefore directionally correct but incomplete, since it treats "safer" as synonymous with "lower pH" while ignoring that dilution changes concentration, not the acid's intrinsic strength.

Marker's note: top-band answers (1) correctly distinguish concentration from strength, (2) quantify the pH change for both a strong and a weak acid using the $1$ unit / $0.5$ unit rules, (3) explicitly use Le Chatelier to explain why the weak-acid change is smaller, and (4) reach an explicit, qualified judgement rather than a flat "true" or "false".